JEE Exam > JEE Questions > Let P be a point on the parabola x2= 4y. If t...
Start Learning for Free
Let P be a point on the parabola x2 = 4y. If the distance of P from the centre of the circle x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P is
- a)x + y + 1 = 0
- b)x + 4y - 2 = 0
- c)x + 2y = 0
- d)x - y + 3 = 0
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
Let P be a point on the parabola x2= 4y. If the distance of P from the...
Let P = (2t, t2)
The equation normal at P to x2 = 4y:
y - t2 = -1/t (x - 2t)
It passes through (-3, 0).
0 - t2 = -1/t (-3 - 2t)
t3 + 2t + 3 = 0
(t + 1)(t2 - t + 3) = 0
⇒ t = -1
Point P is (-2, 1).
The equation of tangent to x2 = 4y at (x', y') is:
xx' = 2(y + y')
So, at P(-2, 1),
x(-2) = 2(y + 1)
x + y + 1 = 0
For the distance between point P and centre of circle to be minimum, line drawn from the centre of circle to point P must be normal to the parabola at P.
The equation normal at P to x2 = 4y:
y - t2 = -1/t (x - 2t)
It passes through (-3, 0).
0 - t2 = -1/t (-3 - 2t)
t3 + 2t + 3 = 0
(t + 1)(t2 - t + 3) = 0
⇒ t = -1
Point P is (-2, 1).
The equation of tangent to x2 = 4y at (x', y') is:
xx' = 2(y + y')
So, at P(-2, 1),
x(-2) = 2(y + 1)
x + y + 1 = 0
Free Test
FREE
| Start Free Test |
Community Answer
Let P be a point on the parabola x2= 4y. If the distance of P from the...
Objective: To find the equation of the tangent to the parabola at point P that is at minimum distance from the center of the circle.
Given:
- Parabola equation: x^2 = 4y
- Circle equation: x^2 + y^2 - 6x - 8 = 0
Approach:
1. Find the coordinates of the center of the circle.
2. Find the distance between the center of the circle and the point P on the parabola.
3. Find the derivative of the parabola equation.
4. Use the derivative to find the slope of the tangent at point P.
5. Use the slope and the coordinates of point P to find the equation of the tangent.
Explanation:
Step 1: Finding the center of the circle
1. Rearrange the equation of the circle to the standard form: x^2 - 6x + y^2 - 8 = 0
2. Complete the square for the x terms: (x^2 - 6x + 9) + y^2 - 8 = 9
3. Simplify: (x - 3)^2 + y^2 = 17
4. The center of the circle is (3, 0).
Step 2: Finding the distance between the center of the circle and point P
1. Let P be the point (x, y) on the parabola.
2. Use the distance formula: distance = sqrt((x - 3)^2 + (y - 0)^2)
3. Simplify: distance = sqrt((x - 3)^2 + y^2)
Step 3: Finding the derivative of the parabola equation
1. Differentiate the parabola equation with respect to x: d/dx (x^2) = d/dx (4y)
2. Simplify: 2x = 4(dy/dx)
3. Rearrange the equation to find dy/dx: dy/dx = x/2
Step 4: Finding the slope of the tangent at point P
1. Substitute x from the coordinates of point P into the derivative: dy/dx = x/2
2. Substitute the x-coordinate of point P from the parabola equation: dy/dx = x/2 = (x^2)/2
3. Simplify: dy/dx = x^2/2
Step 5: Finding the equation of the tangent
1. The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
2. Substitute the slope dy/dx and the coordinates (x, y) of point P into the equation: y - y1 = x^2/2 (x - x1)
3. Simplify: y - y1 = (x^3 - x^2x1)/2
4. Substitute the x-coordinate of point P from the parabola equation: y - y1 = (x^3 - x^2x1)/2 = (x^3 - x^2(
Given:
- Parabola equation: x^2 = 4y
- Circle equation: x^2 + y^2 - 6x - 8 = 0
Approach:
1. Find the coordinates of the center of the circle.
2. Find the distance between the center of the circle and the point P on the parabola.
3. Find the derivative of the parabola equation.
4. Use the derivative to find the slope of the tangent at point P.
5. Use the slope and the coordinates of point P to find the equation of the tangent.
Explanation:
Step 1: Finding the center of the circle
1. Rearrange the equation of the circle to the standard form: x^2 - 6x + y^2 - 8 = 0
2. Complete the square for the x terms: (x^2 - 6x + 9) + y^2 - 8 = 9
3. Simplify: (x - 3)^2 + y^2 = 17
4. The center of the circle is (3, 0).
Step 2: Finding the distance between the center of the circle and point P
1. Let P be the point (x, y) on the parabola.
2. Use the distance formula: distance = sqrt((x - 3)^2 + (y - 0)^2)
3. Simplify: distance = sqrt((x - 3)^2 + y^2)
Step 3: Finding the derivative of the parabola equation
1. Differentiate the parabola equation with respect to x: d/dx (x^2) = d/dx (4y)
2. Simplify: 2x = 4(dy/dx)
3. Rearrange the equation to find dy/dx: dy/dx = x/2
Step 4: Finding the slope of the tangent at point P
1. Substitute x from the coordinates of point P into the derivative: dy/dx = x/2
2. Substitute the x-coordinate of point P from the parabola equation: dy/dx = x/2 = (x^2)/2
3. Simplify: dy/dx = x^2/2
Step 5: Finding the equation of the tangent
1. The equation of a line in point-slope form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
2. Substitute the slope dy/dx and the coordinates (x, y) of point P into the equation: y - y1 = x^2/2 (x - x1)
3. Simplify: y - y1 = (x^3 - x^2x1)/2
4. Substitute the x-coordinate of point P from the parabola equation: y - y1 = (x^3 - x^2x1)/2 = (x^3 - x^2(
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer?
Question Description
Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer?.
Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let P be a point on the parabola x2= 4y. If the distance of P from the centre of the circle x2+ y2+ 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P isa)x + y + 1 = 0b)x + 4y - 2 = 0c)x + 2y = 0d)x - y + 3 = 0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup